JEE Main 2026 (Online) 6th April Evening ShiftQuestion 1
Let $A=\begin{bmatrix}1&0&0\\3&1&0\\9&3&1\end{bmatrix}$ and $B=[b_{ij}]$, $1\le i,j\le3$. If $B=A^{99}-I$, then the value of $\frac{b_{31}-b_{21}}{b_{32}}$ is :
Matrices
Solution
$A=I+N$ with $N=\begin{bmatrix}0&0&0\\3&0&0\\9&3&0\end{bmatrix}$, $N^2$ has only $(3,1)$ entry $9$, $N^3=0$. $A^{n}=I+nN+\binom n2N^2$, so $b_{21}=3n$, $b_{32}=3n$, $b_{31}=9n+9\binom n2$. For $n=99$: $b_{21}=b_{32}=297$, $b_{31}=891+9\cdot4851=44550$. $\frac{44550-297}{297}=149$. Option (C).